Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		double p[] = new double[N];
		for(int i=0;i<N;i++) p[i]=sc.nextDouble();
		sc.close();
		
		double dp[][] = new double[N+1][N+1]; //i枚目を投げた段階の、表の数kの確率
		dp[0][0]=1;//１枚も投げていないとき、表０枚の確率が1
		//配るDP
		for(int i=0;i<N;i++) {
			for(int j=0;j<N;j++) {
				dp[i+1][0] = dp[i][0]*(1-p[i]);
				dp[i+1][j+1] += dp[i][j+1]*(1-p[i]) + dp[i][j]*p[i];
			}
		}
		//N枚投げたあと
		double ans = 0;
		int half = (N+1)/2;
		for(int i=half;i<N+1;i++) {
			ans += dp[N][i];
		}
		System.out.println(ans);
	}
}